A disk has 8 equidistant tracks. The diameters of the innermost and outermost…
2005
A disk has 8 equidistant tracks. The diameters of the innermost and outermost tracks are 1 cm and 8 cm respectively. The innermost track has a storage capacity of 10 MB. What is the total amount of data that can be stored on the disk if it is used with a drive that rotates it with (i) Constant Linear Velocity (ii) Constant Angular Velocity?
Answer: D. (i) 360 MB (ii) 80 MB — Key idea: For constant linear velocity (CLV) the data density along the track (bits per unit length) is constant, so capacity per track is proportional to the…
- A.
(i) 80 MB (ii) 2040 MB
- B.
(i) 2040 MB (ii) 80 MB
- C.
(i) 80 MB (ii) 360 MB
- D.
(i) 360 MB (ii) 80 MB
Attempted by 129 students.
Show answer & explanation
Correct answer: D
Key idea: For constant linear velocity (CLV) the data density along the track (bits per unit length) is constant, so capacity per track is proportional to the track circumference (and hence to track diameter). For constant angular velocity (CAV) with a fixed data rate per revolution, each track stores the same amount of data per revolution, so each track holds the same capacity.
Given: 8 equidistant tracks with diameters 1, 2, 3, 4, 5, 6, 7, 8 cm. Innermost (d = 1 cm) capacity = 10 MB.
CLV calculation:
Track with diameter 1 cm: 10 MB
Track with diameter 2 cm: 20 MB
Track with diameter 3 cm: 30 MB
Track with diameter 4 cm: 40 MB
Track with diameter 5 cm: 50 MB
Track with diameter 6 cm: 60 MB
Track with diameter 7 cm: 70 MB
Track with diameter 8 cm: 80 MB
Sum for CLV = 10 × (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 10 × 36 = 360 MB
CAV calculation:
With constant angular velocity and a fixed data rate per revolution, every track stores the same 10 MB (the innermost track's capacity). Total for CAV = 8 × 10 MB = 80 MB
Final answer: (i) CLV → 360 MB; (ii) CAV → 80 MB.